Physical Sciences Grade 12 Paper 2 KZN Mock 2025 Memo

} KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA GRADE 12 NATIONAL SENIOR CERTIFICATE PHYSICAL SCIENCES P2 (CHEMISTRY) PREPARATORY EXAMINATION SEPTEMBER 2025 MARKING GUIDELINES MARKS: 150 This marking guideline consists of 11 pages. Copyright reserved Please turn over

Physical Sciences/P2 QUESTION 1 1.1 Byv 1.2 Dvv 1.3 Cvv 1.4 Cvv 1:5 Cvv 1.6 Cvv 1.7 Dvv 1.8 Cvv 1.9 D vv DO NOT MARK 1.10 Byv QUESTION 2 2.1.1 4 -bromo—5 — ethyl — 2 - methyiheptanev vv Marking criteria: e correct stem i.e. heptane e substituents correctly identified i.e. bromo, ethyl, methylv e IUPAC name completely correct including numbering, sequence and hyphen ¥ 2.1.2 3-methylbut-1-enevv Marking criteria: ¢ correct stem and substituents i.e. dibromo, methyl and heptanev e {UPAC name completely correct including numbering, sequence and hyphen ~ 22 CrH2nO2v 2.3.1 Ev 232 Cy Copyright reserved September 2025 Preparatory Examination (2) (2) (2) (2) (2) (2) (2) (2) (2) (2) [20] (3) (2) (1) (1) (1)

Physical Sciences/P2 September 2025 Preparatory Examination 2.4.1. Compounds with the same molecular formula, “ but different functional groups/homologous series.” Marking criteria: If any one of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. 24.2 CsHisO2v“v 2.4.3. Pentanoic acid or methylbutanoic acidv v Marking criteria: e correct homologous series: carboxylic acid. ¥ e IUPAC name completely correct. v QUESTION 3 3.1 The temperature at which the vapour pressure of a substance equals atmospheric pressure. v” Marking criteria: If any one of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. The underlined phrases must be in the correct context. $24 Ensure a fair test/comparison. Vv 3.2.2 London forces/Dispersion forces. ¥ 3.2.3. Hydrogen bonding “ 3.4 Ye Y has the lowest boiling point. ” Or Y has the lowest intermolecular forces compared to W and X. 3.5 Marking criteria: e Relate boiling point with length of carbon chain/branching/number of side chains/surface area. VV e Compare the strength of the intermolecular forces. V e¢ Compare the energy required to overcome the intermolecular forces. ¥ X has the lowest boiling point Y and therefore has the shorter carbon chain/is branched/more compact/more spherical/smaller surface area over which the intermolecular forces act than W.v \Wlaakear intarmolaciilar farcac/\/an dar \Alaale farcac/| andan farcac than \Vi 4% (2) (2) (2) [14] (2) (1) (1) (1)

Physical Sciences/P2 September 2025 Preparatory Examination W has the higher boiling point Vv and therefore has the longer carbon chain/no branches/less_compact/less_spherical/largest_surface_area_over which the intermolecular forces act than X.v Stronger/more_ intermolecular _forces/Van_ der Waals forces/London_ forces than Xv More energy needed to overcome the intermolecular forces” 3.6 i H— C — H H | H | | H — f — Cc — i — O-—-H i | 4 H — C —H | H Marking criteria: e Functional group on first carbonY e 3 carbons in the longest chain “ e 2-—methyl groups on the second carbon ~ (3) 3.7 Marking criteria: e LOWER THAN Y ¢ Correctly identify intermolecular forces in both compounds. ~ e¢ Compare the strength of the intermolecular forces. “ e¢ Compare the energy required to overcome the intermolecular forces. ¥ LOWER THANY 2,2 — dimethylpropane has only London/Dispersion forces between molecules and Compound Y has hydrogen bonds” and dipole-dipole forces in addition to London/Dispersion forces Intermolecular forces between molecules of 2,2 — dimethylpropane are weakerY Require less energy to overcome. ” (4) [18] Copyright reserved Please turn over

Physical Sciences/P2 September 2025 Preparatory Examination QUESTION 4 4.1 additiony (1) 4.2 The C-atom bonded to the hydroxyl/-OHY is bonded to three other carbon 4.3.1 4.3.2 4.4.1 4.4.2 4.5.1 4.5.2 4.5.3 atoms.v Marking criteria: If any one of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. The underlined phrases must be in the correct context. (2) Water/H20v (1) Sulphuric acid/H2SOs or Phosphoric acid/HsPO4” (1) H | H — C — H H H H H | | | | H—C—C —¢G—C€C— C —H | | | | | H H H 7 H H Marking criteria: e Correct functional groupY e substituent correctly identified i.e. methylv e IUPAC name completely correct including numbering, sequence and hyphen ¥ (3) 2 - methyl — 1 - bromopentaney vv Marking criteria: ¢ correct stem i.e. pentane « substituents correctly identified i.e. bromo, methyl” « IUPAC name completely correct including numbering, sequence and hyphen ¥ (3) hydroxyl” (1) substitution’ (1) Dilute KOH/potassium hydroxide” (1) Copyright reserved Please turn over

Physical Sciences/P2 September 2025 Preparatory Examination 4.5.4 CeHi3Br + KOH -—> CoH140 + KBr Marking criteria: e Both reactants correct. Vv ¢ Both products correctv e Balancing Vv (3) [17] QUESTION 5 5.1 The COz produced escapes from the container. v (1) 5.2 e Change in concentration V of products/reactants per (unit) time. W e Change in amount/number of moles/volume/mass VY of products/reactants per (unit) time. Vv e Amount/number of moles/volume/mass of products formed/reactants used per (unit) time. vv Marking criteria: If any one of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. The underlined phrases must be in the correct context. e Rate of change in concentration/amount/number of moles/volume/ mass. vv (2 or 0) (2) 5.3 e Reaction is complete/no more COz2 is produced/CaCOs is used upY ¥ (2) 5.4 e Crush the calcium carbonate chips to a powder. e Add a catalyst. v (2) 5:5 Marking criteria: e Correct substitution (= ) in the formula n =a to calculate n(CaCOs) v e Ratio: n(CaCOs) used equals n(COz) produced v e Usen =a to calculate m(COz) v e Substitute for both n and M correctly i.e.: n = 0,6225 and M = 44v e Rate formula: rate = = v e Correct substitution of 1,37 for ratev ¢ Correct substitution for Am in the formula: rate = =v e Final answer At = 19,99 minutesY Copyright reserved Please turn over

Physical Sciences/P2 September 2025 Preparatory Examination n(COz) lost = n(CaCOs) reacted Vv = m 7 M - 62,25 oy 100 = 0,6225 mols m(CO>?) lost = nM = (0,6225)(44) v = 27,399 Am = aoe v rate “At _ 27,39 V 1,37V = At At = 19,99 minutesY (8) 5.6.1. INCREASES.Y (1) 5.6.2 REMAINS THE SAMEY CaCOs is the limiting reagent” (2) 5.7 e More reacting molecules per unit volume. V More molecules correctly orientated. ¥ More effective collisions per unit time/second. ¥ OR e Frequency of effective collisions increases. (3) [21] QUESTION 6 6.1 Add more reactants/increase the amount/mols/concentration of the reactants. v OR Remove some of the products/ decrease the amount/mols/concentration of the products. OR Decrease in pressure (1) Copyright reserved Please turn over

Physical Sciences/P2 6.2 6.3 6.4 Marking criteria: e Correctly calculate 70,59% of 5 = 3,53v « Using the correct mol ratiov e Calculating the quantity(mol) at equilibrium of all three substances Vv e Using the concentration of Q2 at equilibrium and the number of mols of Az at equilibrium in the equation c = n/Vto calculate the volume of the container. vv * Calculating the equilibrium concentrations of the reactantsv e Ke expression’ e Correct substitution of equilibrium concentrations into Ke expression v e Kce= 5,09 v TRQT A @ Initial quantity (mol) 5 0 0 Change (mol) 3,53Y (3,53 1,765) | v Quantity at equilibrium (mol) (1.47 | 3,83 | 1,765>] v Equilibrium concentration (mol.dm-) 0,736 4,765) 0,8825 | v ° " <|5 0,8825 Ke = [Az][Q2 [A2Q? (1,765)2(0,8825) = (0,735)? = 5,09 v The rates of the forward and reverse reaction do not increase equally. v Marking criteria: If any one of the underlined key words/phrases in the correct context is omitted, deduct 1 mark. The underlined phrase must be in the correct context. When the equilibrium in a closed system is disturbed, the system will September 2025 Preparatory Examination re-instate_a_ new equilibrium by favouring the reaction that will oppose the disturbance. vv (2) Copyright reserved Please turn over

Physical Sciences/P2 September 2025 Preparatory Examination 6.5 GREATER THAN v¥ 6.6 The rate of the forward reaction decreases less than the rate of the reverse reaction. ~ The forward reaction is favoured. ¥ Concentration of the products increases while the concentration of the reactants decreases. V QUESTION 7 7.1.1. CHsCOOVY 7.1.2 GREATER THAN 7Y 7.1.3. CH3sCOO- + HzO — CH3sCOOH + OH" LHSvY RHSY Balancingv 2a It is a substance_that_ionises_ completely v in water to produce a high concentration of hydronium ions. ~ e Calculate n(HCfyexcess ¥ e Substitute for c and V inn =cV v e Ratio HCf : NazCOs = 2:1” e Substitute for c and V v e Calculate n(HCf)ota by adding above 2 valuesY e Ratio of c(HC2) = c(Hs0*)” e Substitute into c = n/V to calculate concentration of H30*V e Equation: pH = - log[H30*]“ e Substitute into pH = - log[Hs0*) e Final answer” Copyright reserved Please turn over (1) (3) [17] (1) (1) (3) (2)

Physical Sciences/P2 n(HC£)excess n(HCé) reacted with NazCO3 n ( HC£)totai c(H30*) pH wou od September 2025 Preparatory Examination cV v (0,013)(0,075) v 0,000975 mols 2cV v (2)(0,075)(0,025) 0,00375 mols 0,000975 + 0,00375v 0,004725 mols (HC?) ¥ n Vv 0,004725 0,05 0,0945 mol.dm*? -log[H3O*]v -log0,0945v 1,02” 2H* + 2e =He (%) QUESTION 8 8.1 Pty 8.2 Remain the same. Vv 8.3.1 H2—>2H* + 2e vv Notes « Ignore phases © Hee-2H? + 2e (9%) H22H* + 2e (4) Ignore if charge on electron omitted. 8.3.2 25°C/298 K v 8.3.3. 1 mol.dm? v 8.4 Cutt v Copyright reserved Please turn over (10) [17] (1) (1) (2) (1) (1) (1)

Physical Sciences/P2 September 2025 Preparatory Examination 8.5 Pt(s) /H2(g)/H*(aq)(1 mol.dm-?) v // ¥ Cu2*(aq)(1_ mol.dm?)/Cu(s) ¥ (3) Ignore the phases and concentrations 8.6 Notes « Accept any other correct formula from the data sheet. Any other formula using unconventional abbreviations, e.g. E° cei = E°oa- E°ra followed by correct substitutions Max: y, EX cei = E* reduction - E*oxidation ¥ 0,75 v = 0,34v - E*oxidation E°oxidation = -0,41VVY Yesv, Cr?* is not a solid. ¥ (6) [16] QUESTION 9 9.1 Electrolyticv They both have a power supply” Both cells converts electrical energy to chemical energy. (2) 9.2 An electrolyte is a substance of which the aqueous solution contains ions. vv OR A substance that dissofves in water to give a solution that conducts electricity. (2) 9.3 2H20 + 2e — H2 + 20H YY Ignore phases Notes e 2H20+2e5H2+20H = (1/2) e H2+2OH"< 2H20+2e (2/2) e H2+20OH"S 2H20+2e (0/2) e 2H20 + 2e°<H2+20H° (0/2) (2) 9.4 Cu or CopperY (1) 9.5 DecreasesY Cu** (or Copper(Il) ions) are reducedv to Cu v (or Copper). (3) [10] TOTAL: 150 NB: MARK SCRIPT OUT OF 148 AND THEREAFTER CONVERT TO TOTAL MARKS : 150 Copyright reserved